Question

what is csc(∠a)? reduce fractional answers to lowest terms.
(image of right triangle with right angle at c, ac = 5, bc = 12, ab = 13, vertices labeled a, b, c)

Explanation:

Step1: Recall the definition of cosecant

The cosecant of an angle in a right triangle is the reciprocal of the sine of that angle. For an angle \( \theta \), \( \csc(\theta)=\frac{1}{\sin(\theta)} \), and \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \), so \( \csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}} \).

Step2: Identify the sides relative to \( \angle A \)

In right triangle \( ABC \) with right angle at \( C \), for \( \angle A \):

• The opposite side to \( \angle A \) is \( BC = 12 \).
• The hypotenuse of the triangle is \( AB = 13 \) (the side opposite the right angle).

Step3: Calculate \( \csc(\angle A) \)

Using the formula for cosecant, \( \csc(\angle A)=\frac{\text{hypotenuse}}{\text{opposite to } \angle A}=\frac{AB}{BC}=\frac{13}{12} \)? Wait, no, wait. Wait, let's re - check the sides. Wait, angle \( A \): the sides: \( AC = 5 \), \( BC = 12 \), \( AB = 13 \). Wait, for angle \( A \), the opposite side is \( BC = 12 \)? No, wait, in triangle \( ABC \), angle at \( A \), so the sides: adjacent side to \( A \) is \( AC = 5 \), opposite side is \( BC = 12 \), hypotenuse is \( AB = 13 \). Wait, but \( \sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{12}{13} \), so \( \csc(A)=\frac{1}{\sin(A)}=\frac{13}{12} \)? Wait, no, wait, maybe I mixed up the opposite side. Wait, angle \( A \): the right angle is at \( C \), so the sides: \( AC \) and \( BC \) are the legs, \( AB \) is the hypotenuse. So for angle \( A \), the opposite side is \( BC \) (since it's opposite angle \( A \)), and the adjacent side is \( AC \). So \( \sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{12}{13} \), so \( \csc(A)=\frac{13}{12} \)? Wait, but let's check again. Wait, maybe I made a mistake in identifying the opposite side. Wait, angle \( A \): the vertex is at \( A \), so the sides: from \( A \), the two sides are \( AC \) (length 5) and \( AB \) (hypotenuse, length 13), and the side opposite \( A \) is \( BC \) (length 12). So yes, \( \sin(A)=\frac{BC}{AB}=\frac{12}{13} \), so \( \csc(A)=\frac{13}{12} \). Wait, but wait, another way: \( \csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}} \), so hypotenuse is 13, opposite to \( A \) is 12, so \( \csc(A)=\frac{13}{12} \).

Wait, no, wait, I think I messed up. Wait, angle \( A \): the sides: \( AC = 5 \), \( BC = 12 \), \( AB = 13 \). Let's use SOHCAHTOA. Sine is opposite over hypotenuse. For angle \( A \), the opposite side is \( BC = 12 \), hypotenuse is \( AB = 13 \), so \( \sin(A)=\frac{12}{13} \), so \( \csc(A)=\frac{13}{12} \).

Answer:

\( \frac{13}{12} \)